The tungsten filament of a light bulb has a 0.100 millimeter diameter. The filament has a resistance of 0.20 ohms at a temperature of 20 degrees Celsius. Find the length of the filament. Use a value of 5.60 times 10 to the negative eighth ohm meters for the resistivity of tungsten at 20.0 degrees Celsius.
We can call the length of the filament we want to solve for capital 𝐿. As we get started on our solution, we can recall a relationship between resistance, capital 𝑅, and resistivity represented by the Greek letter 𝜌. Given some amount of a material, we could measure the ability of that material to resist current flow. That would be a measure of its resistivity. The resistivity that we would measure would depend on the cross-sectional area of the object as well as how long it was.
We can imagine though that it might be useful to know the electrical resistance of that material by itself, regardless of its cross-sectional area and length. Its that resistance that’s represented by the letter capital 𝑅. This is a measure of how much a material resists the flow of current with all the geometry of the material dropped out. These two terms are mathematically related. If we multiply a given wire’s resistivity by the length of that wire divided by its cross-sectional area, then we find its resistance.
In our particular scenario, we want to solve for the length of our wire. That’s represented by capital 𝐿. So we can use this relationship to solve for it. The length of our filament 𝐿 is equal to the resistance of the wire multiplied by its cross-sectional area over its resistivity. In our problem statement, we’re told the resistance that the filament has at a given temperature as well as its resistivity at that temperature. And we’re also told the diameter of the wire in units of millimeters.
When we plug these values in to our expressions to solve for 𝐿, we use the fact that the cross-sectional area of the circle, in terms of its diameter, is 𝜋 times 𝑑 squared over four. And when we plug in the diameter, we’re careful to convert it into units of meters, so that its units are consistent with those in the rest of the expression.
Considering the units in our overall expression, we see that ohms cancels out of the numerator and denominator, as does one factor of meters. Leaving us with an answer in units of meters which we would expect for a length. Calculating 𝐿 to two significant figures, we find a result of 28 times 10 to the negative third meters or 28 millimeters. That’s the length of the filament of this light bulb.